Relativistic Paradoxes Part B

From the sci.physics FAQ

The Twin Paradox

updated 17-AUG-1992 by SIC

original by Kurt Sonnenmoser

A Short Story about Space Travel:

Two twins, conveniently named A and B, both know the rules of Special Relativity. One of them, B, decides to travel out into space with a velocity near the speed of light for a time T, after which she returns to Earth. Meanwhile, her boring sister A sits at home posting to Usenet all day. When A finally comes home, what do the two sisters find? Special Relativity (SR) tells A that time was slowed down for the relativistic sister, B, so that upon her return to Earth, she knows that B will be younger than she is, which she suspects was the the ulterior motive of the trip from the start.

But B sees things differently. She took the trip just to get away from the conspiracy theorists on Usenet, knowing full well that from her point of view, sitting in the spaceship, it would be her sister, A, who was travelling ultrarelativistically for the whole time, so that she would arrive home to find that A was much younger than she was. Unfortunate, but worth it just to get away for a while.

What are we to conclude? Which twin is really younger? How can SR give two answers to the same question? How do we avoid this apparent paradox? Maybe twinning is not allowed in SR? Read on.

Paradox Resolved:

Much of the confusion surrounding the so-called Twin Paradox originates from the attempts to put the two twins into different frames --- without the useful concept of the proper time of a moving body.

SR offers a conceptually very clear treatment of this problem. First chose one specific inertial frame of reference; let's call it S. Second define the paths that A and B take, their so-called world lines. As an example, take (ct,0,0,0) as representing the world line of A, and (ct,f(t),0,0) as representing the world line of B (assuming that the the rest frame of the Earth was inertial). The meaning of the above notation is that at time t, A is at the spatial location (x1,x2,x3)=(0,0,0) and B is at (x1,x2,x3)=(f(t),0,0) --- always with respect to S.

Let us now assume that A and B are at the same place at the time t1 and again at a later time t2, and that they both carry high-quality clocks which indicate zero at time t1. High quality in this context means that the precision of the clock is independent of acceleration. [In principle, a bunch of muons provides such a device (unit of time: half-life of their decay).]

The correct expression for the time T such a clock will indicate at time t2 is the following [the second form is slightly less general than the first, but it's the good one for actual calculations]:

	    t2          t2      _______________ 
	    /           /      /             2 |
      T  =  | d\tau  =  | dt \/  1 - [v(t)/c]              (1)
	    /           /
	  t1          t1

where d\tau is the so-called proper-time interval, defined by

	      2         2      2      2      2
     (c d\tau)  = (c dt)  - dx1  - dx2  - dx3 .


		   d                          d
	   v(t) = -- (x1(t), x2(t), x3(t)) = -- x(t)
		  dt                         dt

is the velocity vector of the moving object. The physical interpretation of the proper-time interval, namely that it is the amount the clock time will advance if the clock moves by dx during dt, arises from considering the inertial frame in which the clock is at rest at time t --- its so-called momentary rest frame (see the literature cited below). [Notice that this argument is only of a heuristic value, since one has to assume that the absolute value of the acceleration has no effect. The ultimate justification of this interpretation must come from experiment.]

The integral in (1) can be difficult to evaluate, but certain important facts are immediately obvious. If the object is at rest with respect to S, one trivially obtains T = t2-t1. In all other cases, T must be strictly smaller than t2-t1, since the integrand is always less than or equal to unity. Conclusion: the traveling twin is younger. Furthermore, if she moves with constant velocity v most of the time (periods of acceleration short compared to the duration of the whole trip), T will approximately be given by

			      /          2 | 
		    (t2-t1) \/  1 - [v/c]    .             (2)

The last expression is exact for a round trip (e.g. a circle) with constant velocity v. [At the times t1 and t2, twin B flies past twin A and they compare their clocks.]

Now the big deal with SR, in the present context, is that T (or d\tau, respectively) is a so-called Lorentz scalar. In other words, its value does not depend on the choice of S. If we Lorentz transform the coordinates of the world lines of the twins to another inertial frame S', we will get the same result for T in S' as in S. This is a mathematical fact. It shows that the situation of the traveling twins cannot possibly lead to a paradox within the framework of SR. It could at most be in conflict with experimental results, which is also not the case.

Of course the situation of the two twins is not symmetric, although one might be tempted by expression (2) to think the opposite. Twin A is at rest in one and the same inertial frame for all times, whereas twin B is not. [Formula (1) does not hold in an accelerated frame.] This breaks the apparent symmetry of the two situations, and provides the clearest nonmathematical hint that one twin will in fact be younger than the other at the end of the trip. To figure out which twin is the younger one, use the formulae above in a frame in which they are valid, and you will find that B is in fact younger, despite her expectations.

It is sometimes claimed that one has to resort to General Relativity in order to "resolve" the Twin "Paradox". This is not true. In flat, or nearly flat space-time (no strong gravity), SR is completely sufficient, and it has also no problem with world lines corresponding to accelerated motion.


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